Streamline BC on surfaces
 

Hi there,

I need your help on the following question:

I have got the velocity field for the flow past a few randomly placed cylinders (in 2D). I want draw the streamlines by solving the Poisson equation. The BC for the inlet is easy, simply integrating the velocity. However, I do not know how to set the BC on the cylinder surfaces. I will be thankful if somebody can show me how.

Thanks

Frank

Re: Streamline BC on surfaces
 

By convention one sets the streamfunction to zero on solid surfaces. This ensures that streamlines (iso-lines of the streamfunction) are locally parallel to the solid surface.

Re: Streamline BC on surfaces
 

Thanks for your reply. Should I set zero to all surfaces? Streamfunctions is constant on solid surfaces but might be different values. I just do not know how to get these values. Any idea?

Re: Streamline BC on surfaces
 

The stream function is constant on each no-flow surface. If there is a single, connected surface such as the lid-driven cavity problem, there is one constant which can be chosen to be zero. If there are two surfaces such as in duct flow, the stream function is constant on each surface, but the difference in the constants is equal to the flow through the duct, as you remarked.

If there is a single cylinder, the stream function value on the cylinder determines the fraction of flow to the left vs. the right (it could even be greater (or less) than either boundary surface if the fluid was swirling around the cylinder such as would be the case if the cylinder were spinning).

For your problem, I would assign some convenient value at one surface, then evaluate a line integral of the velocity component perpendicular to the line from the initial surface to each of the other (cylindrical)surfaces, just as you would do at the inlet, to evaluate the constant for that surface.

Re: Streamline BC on surfaces
 

Hi Jonas,

Thanks for your reply. The values of streamline function on the cylinder surfaces are constant. But are they time dependant? Since some of the stagnation points on the cylinders may vary with time in my case, say, one cylinder is placed in the wake of another one.

I can prescribe values at the inlet, find the stagnation points on the cylinders and trace back their characteristic roots at the inlet. But this procedure is very time-consuming and has to be done at every time step. I am wondering if there is any more efficient way.

Thanks again.

Re: Streamline BC on surfaces
 

Frank said: The values of streamline function on the cylinder surfaces are constant. But are they time dependant? Since some of the stagnation points on the cylinders may vary with time in my case, say, one cylinder is placed in the wake of another one.

As you suggested, the stream function on the cylinders may be a function of time, especially in the case of a cylinder in the wake of another. Using the procedure of recovering the stream function from the flow field, perhaps you could assume values, do a fit, and then look for new values which would reduce the error in the fit by a search. One might vary the values and look at the change in fit/error, then choose new values based on the differences.

Yes, there may be a better way, depending on how the flow field was computed. Using the finite element method, the problem might be solved using a particular type of Hermite element, which has the stream function and velocity components (dPsi/dy and -dPsi/dx) as degrees of freedom. The velocity elements are the curl of a stream function element. The stream function can be found for visualization by interpolating the solution with the stream function element.

The form of the element on a rectangle can be found in a classic test by Zienkiewicz disguised as a plate element (you have to fix up the normalization). The derivative degrees of freedom must be interchanged and a sign changed (d/dx, d/dy -> d/dy, -d/dx). Then take the curl of the element to get the velocity element. The velocity element is divergence-free and has sufficient continuity that the divergence vanishes on the boundary as well.

I haven't explained how to transform to a general quadrilateral, but this works for me.

Jonas